Tensor Decomposition

Solution 1:

If $V$ would have been a vector space over $\mathbb C$ instead, there is only one value of $R$ where the set of tensors having rank $R$ has non-zero (Lebesgue) measure (this single value of $R$ is called the generic rank). As Yrogirg says, this $R$ is expected to be $$\left\lceil \frac{m^n}{mn - m + 1}\right\rceil.$$ However, this is not always the case. For example, over $\mathbb C^3 \otimes \mathbb C^3 \otimes \mathbb C^3$ the generic rank is 5.

Over $\mathbb R$ the situation is more complicated and we can have multiple values of $R$ where the set of tensors having rank $R$ has non-zero measure. These $R$ are called typical ranks. For example, in $\mathbb R^2 \otimes \mathbb R^2 \otimes \mathbb R^2$ both 2 and 3 are typical ranks (and 3 is the maximal rank). Of course, in the case of $\mathbb R^n \otimes \mathbb R^n$ the single typical rank is $n$.

Determining the typical ranks for tensors over $\mathbb R$ is an open question, and it's mostly third order tensors which have been studied. A way of determining the minimal typical rank over $\mathbb R$ numerically is described in P. Comon, J.M.F. ten Berge, L. De Lathauwer, J.Castaing (2009), Generic and typical ranks of multi-way arrays, Linear Algebra and its Applications, 430, 2997-3007.

Solution 2:

Well, I've found some sort of conjecture/rule of thumb, that the "expected rank" for both (?) complex and real tensors is almost everywhere $$\frac{m^n}{mn - m + 1}$$ At least that's how I understood it. It seems that this estimation could be obtained rather trivially, just by counting the number of degrees of freedom, but I couldn't understand that $-m+1\;$ part.

Check "Tensor Decompositions, Alternating Least Squares and other Tales." by P. Comon, X. Luciani and A. L. F. de Almeida for the details. In addition, it is a good starting to point for people interested in tensor decomposition and Comon's webpage features software (MATLAB codes) for tensor decomposition.